KEY Statistics Test 3 Name - Tim Busken

Statistics Test 3 Name:KEY

1. Find the area under the standard normal distributioncurve for each.

(a) Between z = −0.19 and z = 1.23.

normalCdf(−0.19, 1.23).= 0.4660

(b) To the left of z = −1.56.

normalCdf(−100, −1.56).= 0.0594

(c) To the right of z = −0.38.

normalCdf(−0.38, 100).= 0.6480

2. Find each probability using the standard normal distribu-tion curve for each.

(a) P (−0.09 < z < 2.42)

= normalCdf(−0.09, 2.42).= 0.5281

(b) P (z > −1.68)

= normalCdf(−1.68, 100).= 0.9535

(c) P (z < 0.23)

= normalCdf(−100, 0.23).= 0.5910

3. Find the indicated z score. The graph depicts the standardnormal distribution with mean 0 and standard deviation 1.

(a)

z = invnorm(0.8980).= 1.27

(b)

z = invnorm(1− 0.2873).= 0.56

4. Find the critical value of z.

(a) z0.12 = invorm(1− 0.12).= 1.17

(b) z0.08 = invorm(1− 0.08).= 1.41

5. Find the critical value of z that represents the 45th percentile.

z = invnorm(0.45).= -0.13

6. The waiting time in line at a Starschmuchs Coffee is normally distributed with amean of 3.2 minutes and a standard deviation of 1.3 minutes. Find the probabilitythat a randomly selected customer has to wait

(a) Less than 1 minute.

µ = 3.2 and σ = 1.3. Let X = the continuous random variable (CRV) repre-senting a randomly selected wait time. We convert the distribution of X valuesover to its distribution of z scores, and find the corresponding area under thestandard(ized) normal curve.

Normal Distribution of Wait Times

Standard Normal Distribution

P (X < 1) = P

(z <

X − µσ

)

= P

(z <

1− 3.2

1.3

).= P (z < −1.69)

= normalCdf(−100, −1.69)

.= 0.0455

X

z

Alternatively, we could just find the area under the distribution of wait times, X, with

P (X < 1) = normalCdf(−∞, X, µ, σ) = normalCdf(−1010, 1, 3.2, 1.3).= 0.0453

(b) more than 2 minutes.

P (X > 2) = P

(z >

X − µσ

)

= P

(z >

2− 3.2

1.3

).= P (z > −0.92)

= normalCdf(−0.92, 100)

.= 0.8212



X

z

Alternatively, we could just find the area under the distribution of wait times, X, with

P (X > 2) = normalCdf(X,∞, µ, σ) = normalCdf(2, 1010, 3.2, 1.3).= 0.8220

(c) between 0.75 minutes and 2 minutes.

P (0.75 < X < 2)

= P

(0.75− 3.2

1.3< z <

2− 3.2

1.3

).= P (−1.88 < z < −0.92)

= normalCdf(−1.88, −0.92)

.= 0.1487



X

z

(P (0.75 < X < 2) = normalCdf(0.75, 2, 3.2, 1.3)

.= 0.1482

)7. The average yearly precipitation in San Diego is 9.62 inches with a standard de-

viation of 4.42 inches and precipitation amounts are normally distributed.

(a) Find the probability that a randomly selected year will have precipitationgreater than 12 inches.

µ = 9.62 in. and σ = 4.42 in. Let X = the continuous random variable (CRV)representing a randomly selected yearly precipitation amount.

P (X > 12) = P

(z >

X − µσ

)

= P

(z >

12− 9.62

4.42

).= P (z > 0.54)

= normalCdf(0.54, 100)

.= 0.2946

Normal Distribution of Precipitations


X

z

Alternatively, we could just find the area under the distribution of precipitations, X, with

P (X > 12) = normalCdf(X,∞, µ, σ) = normalCdf(12, 1010, 9.62, 4.42).= 0.2951

(b) Find the probability that five randomly selected years will have an average precipitationgreater than 8 inches.

Let X = the continuous random variable (CRV) representinga randomly selected sample mean yearly precipitation amount.

P (X > 8) = P

(z >

X − µσ/√n

)

= P

(z >

8− 9.62

4.42/√

5

)X

Sampling Distn.of Sample Means

µX = 9.62 and σX = 4.42/√

5.= 1.98

= P (z > −0.82)

= normalCdf(−0.82, 100)

.= 0.7939


z

Alternatively, we could just find the area under the sampling distribution of sample meanprecipitations, X, with

P (X > 8) = normalCdf(X,∞, µX , σX) = normalCdf(8, 1010, 9.62, 4.42/√

5).= 0.7938

(7c cont.) Find the precipitation amount from the distribution of precipitationsthat represents the 75th percentile.

precipitation, X

X = invnorm(percentile, µ, σ) = invnorm(0.75, 9.62, 4.42).= 12.6 in

Alternatively, we first find the zvalue from the standard normaldistribution that corresponds tothe 75th percentile.

z = invnorm(0.75).= 0.6744897495

Second, we solve z =X − µσ

for X to get

X = µ+ z · σ

Afterwards, replace µ with 9.62, z with0.6744897495 and σ with 4.42 so that

X = 9.62+(0.6744897495)·(4.42).= 12.6 in.

8. Some passengers died when a water taxi sank in Baltimore’s inner harbor. Menare typically heavier than women and children, so when loading a water taxi, let’sassume a worst-case scenario in which all passengers are men. Based on data fromthe National Health and Nutrition Survey, assume that weights of men are normallydistributed with a mean of 172 lb. and a standard deviation of 29 lb.

(a) Find the probability that if an individual man is randomly selected, his weightwill be greater than 175 lb.

µ = 172 lb. and σ = 29 lb. Let X = the continuous random variable (CRV)representing the weight of a randomly selected man.

P (X > 175) = P

(z >

X − µσ

)

= P

(z >

175− 172

29

).= P (z > 0.10)


.= 0.4602


Distn. of Men’s Weights, X

z

Alternatively, we could just find the area under the distribution of weights, X, with

P (X > 175) = normalCdf(X,∞, µ, σ) = normalCdf(175, 1010, 172, 29).= 0.4588

(b) Find the probability that 20 men will have a mean weight that is greater than175 lb. (so that their total weight exceeds the safe capacity of 3500 lb.

Let X = the continuous random variable (CRV)representing a randomly selected sample mean weight.

X


µX = 172 and

σX = 29/√20

.= 6.5

P (X > 175) = P

(z >

X − µσ/√n

)

= P

(z >

175− 172

29/√

20

).= P (z > 0.46)


.= 0.3228

Alternatively, we could just find the area under the sam-pling distribution of sample mean weights, X, with

P (X > 175) = normalCdf(X,∞, µX , σX)

= normalCdf(175, 1010, 172, 29/√

20).= 0.3218

9. The average per capita spending on health care in the United States is $5274.The standard deviation is $600 and the distribution of health care spending is ap-proximately normal. Find the limits of the middle 50% of individual health careexpenditures.

Standard Normal DistributionLet X be the continuous random variable(CRV) representing a randomly selected in-dividual health care expenditure. Find thez scores from the standard normal distri-bution corresponding to the 25th and 75thpercentiles. Then use

X = µ+ z · σ

.z0.75 = invnorm(0.25).= −0.6744897495,

and by symmetry, z0.25.= 0.6744897495

xlow = µ+z·σ .= $5274+(−0.6744897495)·($600) ≈ $4869.31

xhigh = µ+z·σ .= $5274+(0.6744897495)·($600) ≈ $5678.69

Alternatively, we could bypass using the standard normal distribution of z scoreswith:

xlow = invnorm(percentile, µ, σ) = invnorm(0.25, 5274, 600).= $4869.31

xhigh = invnorm(percentile, µ, σ) = invnorm(0.75, 5274, 600).= $5678.69

10. A prestigious college decides to only take applications from student who havescored in the top 5% on the SAT test. The SAT scores are approximately normallydistributed with a mean of 490 and a standard deviation of 70. Find the score thatis necessary to obtain in order to qualify for applying to this college.

Let X be the continuous random variable (CRV) representing SAT scores.

A student must score in the 95thor higher percentile in order tobe admitted. First, we find the zvalue from the standard normaldistribution that corresponds tothe 95th percentile.

z = invnorm(0.95).= 1.644853626

Second, we solve z =X − µσ

for X to get

X = µ+ z · σ

Afterwards, replace µ with 490, z with1.644853626 and σ with 70 so that

X = 490 + (1.644853626) · (70).= 605

An alternate solution route is

X = invnorm(percentile, µ, σ)

= invnorm(0.95, 490, 70).= 605

11. Americans ate an average of 25.7 pounds of Krusty-O Cereal each last year andspent an average of $61.50 per person doing so. If the standard deviation for con-sumption is 3.75 pounds and the standard deviation for the amount spent is $5.89,find the following:

(a) The probability that the sample mean Krusty-O cereal consumption for a ran-dom sample of 40 American consumers exceeded 27 pounds.

Let X = the continuous random variable (CRV) representing arandomly selected sample mean consumption amount.

P (X > 27) = P

(z >

X − µσ/√n

)

= P

(z >

27− 25.7

3.75/√

40

).= P (z > 2.19)


.= 0.0143

X

z


Std. NormalDistn.

µX = 25.7 lb. and

σX = 3.75/√40

.= 0.59 lb.

Alternatively, we could just find the area under the sampling distribution with

P (X > 27) = normalCdf(X,∞, µX , σX) = normalCdf(27, 1010, 25.7, 3.75/√

40).= 0.0142

(b) The probability that for a random sample of 50, the the average yearly amountspent on Krusty-O Cereal was between $60.00 and $100.

Let X = the continuous random variable (CRV) representing a randomly se-lected sample mean yearly amount spent.

P (60 < X < 100)

= P

(X − µσ/√n< z <

X − µσ/√n

)

= P

(60− 61.50

5.89/√

50< z <

100− 61.50

5.89/√

50

).= P (−1.8 < z < 46.22)

= normalCdf(−1.8, 46.22).= 0.9641

X

z


µX = $61.50 and

σX = 5.89/√50

.= $0.83

Std. NormalDistn.

Alternatively, we could just find the area under the sampling distribution with

P (60 < X < 100) = normalCdf(60, 100, 61.50, 5.89/√

50).= 0.9641

12. Use the normal approximation of the binomial probabilitydistribution to find the probabilities for the discrete randomvariable, X.

(a) Find the probability that X is 19, assuming n = 40 and p = 0.5.We find that µ = n · p = 40 · 0.5 = 20, σ =

√npq =

√10, and both np ≥ 5

and nq ≥ 5. Thus,

P (X = 19).= P (18.5 < X < 19.5)

= P

(X − µσ

< z <X − µσ

)

= P

(18.5− 20√

10< z <

19.5− 20√10

).= P (−0.47 < z < −0.16)

= normalCdf(−0.47, −0.16)

.= 0.1173

Binomial Probability Distribution


X

z

(b) Find the probability that X is 3, assuming n = 25 and p = 0.4.

We find that µ = n · p = 25 · 0.4 = 10, σ =√npq =

√6, and both np ≥ 5 and

nq ≥ 5. Thus,

P (X = 3).= P (2.5 < X < 3.5)

= P

(X − µσ

< z <X − µσ

)

P

(2.5− 10√

6< z <

3.5− 10√6

).= P (−3.06 < z < −2.65)

= normalCdf(−3.06, −2.65)

.= 0.0029



X

z

(c) Find the probability that X is at least 15, assuming n = 30 andp = 0.5.


√7.5, and both np ≥ 5

and nq ≥ 5. Thus,



P (X ≥ 15).= P (X > 14.5)

= P

(z >

X − µσ

)

= P

(z >

14.5− 15√7.5

).= P (z > −0.18)

= normalCdf(−0.18, 100)

.= 0.5714

X

z13. Use the normal approximation of the binomial probability

distribution to find the probabilities for the discrete randomvariable, X.

(a) Find the probability that X is fewer than 5, assuming n = 300and p = 0.07.


√19.53, and both

np ≥ 5 and nq ≥ 5. Thus,



X

z

P (X < 5).= P (X < 4.5)

= P

(z <

X − µσ

)

= P

(z <

4.5− 21√19.53

).= P (z < −3.73)

= normalCdf(−100, −3.73)

.= 0.0001

(b) Find the probability that X is at most 8, assuming n = 100 andp = 0.13.


√11.31, and both

np ≥ 5 and nq ≥ 5.Thus,



P (X ≤ 8).= P (X < 8.5)

= P

(z <

X − µσ

)

= P

(z <

8.5− 13√11.31

).= P (z < −1.33)

= normalCdf(−100, −1.33)

.= 0.0918

X

z

(c) Find the probability that X is more than 35, assuming n = 50and p = 0.6.


√12 = 2

√3, and both

np ≥ 5 and nq ≥ 5.Thus, Binomial Probability Distribution


P (X > 35).= P (X > 35.5)

= P

(z >

X − µσ

)

= P

(z >

35.5− 30√12

).= P (z > 1.59)


.= 0.0559

X

z

14. Use the normal approximation of the binomial probability distribution.Two out of five adult smokers acquired the habit by age 14. If 400 smokers arerandomly selected, find the probability that 170 or fewer acquired the habit by age14.

We determine that p =2

5= 0.4 and n = 400. Let X be the discrete random variable

representing the number of smokers out of 400 who acquired the habit by age 14.Then, µ = n · p = 400 · 0.4 = 160, σ =

√npq =

√96 = 4

√6, and both np ≥ 5 and

nq ≥ 5. Thus,Binomial Probability Distribution


P (X ≤ 170).= P (X < 170.5)

= P

(z <

X − µσ

)

= P

(z <

170.5− 160√96

).= P (z < 1.07)

= normalCdf(−100, 1.07)

.= 0.8577

X

z

15. Use the normal approximation of the binomial probability distribution.According to Mars (the candy company), 24% of M&Ms plain candies are blue.Assuming that the claimed blue M&Ms rate of 24% is correct, find the probabilityof randomly selecting 100 M&Ms and getting at most 20 that are blue.

We determine that p = 0.24 and n = 100. Let X be the discrete random variablerepresenting the number of blue M&Ms out of 100. Then, µ = n ·p = 100 ·0.24 = 24,σ =√npq =

√18.24, and both np ≥ 5 and nq ≥ 5. Thus,



P (X ≤ 20).= P (X < 20.5)

= P

(z <

X − µσ

)

= P

(z <

20.5− 24√18.24

).= P (z < −0.82)

= normalCdf(−100, −0.82)

.= 0.2061

X

z

16. Find the critical value zα/2 that corresponds to a 92% confidence in-terval.

A 92% Confidence interval meansthat

0.92 = 1− α,so that α = 0.08. Then,

α/2 = 0.08/2 = 0.04

and

zα/2 = z0.04 = invnorm(1−α/2) = invnorm(0.96).= 1.75

17. First-semester GPAs for a random selection of freshmen at a large uni-versity are shown below. Estimate the true mean GPA of the freshmanclass with 99% confidence. Assume σ = 0.62 and that the distributionof first-semester GPAs is normal.

1.9 3.2 2.0 2.9 2.7 3.32.8 3.0 3.8 2.7 2.0 1.92.5 2.7 2.8 3.2 3.0 3.83.1 2.7 3.5 3.8 3.9 2.7

We find the sample mean is X.= 2.9125 and 99% confidence implies

zα/2 = z0.005 = invnorm(1− 0.005).= 2.58.

Then,

X − zα/2·(σ√n

)< µ < X + zα/2·

(σ√n

),

or

2.9125−2.58

(0.62√

24

)< µ < 2.9125+2.58

(0.62√

24

),

or

2.58 < µ < 3.24

18. Find the critical value tα/2 that corresponds to a 90% interval, assum-ing n = 10.

A 90% Confidence interval meansthat

0.90 = 1− α,so that α = 0.10. Then,

α/2 = 0.10/2 = 0.05

and

tα/2 = t0.05 = invT (1−α/2, n−1) = invT (0.95, 9).= 1.833112 ≈ 1.83

19. The approximate costs (in thousands) for a 30-second spot for variouscable networks in a random selection of cities are shown below. Esti-mate the true population mean cost for a 30-second advertisement oncable network with 90% confidence.Assume the population of costs isapproximately normal.

14 55 165 9 15 66 23 30 15022 12 13 54 73 55 41 78

We find the sample mean is X.= 51.4705, sample standard deviation is

s.= 45.9839 and 90% confidence implies tα/2 = t0.05 = invT (0.95, 16)

.=

1.74588 ≈ 1.75.

Then,

X − tα/2·(

s√n

)< µ < X + tα/2·

(s√n

),

or

51.4705−1.74588

(45.9839√

17

)< µ < 51.4705+1.74588

(45.9839√

17

),

or

32.0 < µ < 70.9

20. A university dean of students wishes to estimate the average numberof hours students spend doing homework per week. The standarddeviation from a previous study is 6.2 hours. How large a samplemust be selected if he wants to be 99% confident of finding whetherthe true mean differs from the sample mean by 1.5 hours?

We are asked to determine the size, n, of asample necessary for an interval estimateof the average weekly study amount. Theformula is

n =

(zα/2 · σE

)2

We are told to assume σ = 6.2 and E =1.5. If 99% Confidence is desired, then0.99 = 1 − α, or α = 0.01. This implieswe should use zα/2 = z0.005 = invnorm(0.995).= 2.58. Then,

n =

(zα/2 · σE

)2

=

((2.58) · (6.2)

1.5

)2.= 114

21. Thirty randomly selected students took the calculus final. If the sam-ple mean was 95 and the standard deviation was 6.6, construct a 99%confidence interval for the mean score of all students.We are not given σ, so we use a t distribution. We are given n = 30, X = 95 ands = 6.6. 99% confidence implies 0.99 = 1− α, or α = 0.01, andtα/2 = t0.005 = invT (0.995, 29)

.= 2.75638

Then,

X − tα/2·(

s√n

)< µ < X + tα/2·

(s√n

),

or

95−2.75638

(6.6√

30

)< µ < 95+2.75638

(6.6√

30

),

or

92 < µ < 98

22. A study of 35 golfers showed that their average score on a particularcourse was 92. The standard deviation of the population is 5. Findthe 95% confidence interval of the mean score for all golfers.We are given σ = 5, so we use a z distribution. We are given n = 35 and X = 92.95% confidence implies 0.95 = 1− α, or α = 0.05, andzα/2 = z0.025 = invnorm(0.975)

.= 1.96

Then,

X − zα/2·(σ√n

)< µ < X + zα/2·

(σ√n

),

or

92− 1.96

(5√35

)< µ < 92 + 1.96

(5√35

),

or

90.3 < µ < 93.7

23. A recent study of 75 workers found that 53 people rode the bus towork each day. Find the 95% confidence interval of the proportion ofall workers who rode the bus to work.

We are given n = 75 and X = 53, so the sample proportion is p̂ =53

75.

95% confidence implies 0.95 = 1− α, or α = 0.05,and zα/2 = z0.025 = invnorm(0.975)

.= 1.96

Then,

p̂− zα/2 ·√p̂ · q̂n

< p < p̂+ zα/2 ·√p̂ · q̂n,

or

53

75−1.96·

√√√√√ 53

75·(

1− 53

75

)75

< p <53

75+1.96·

√√√√√ 53

75·(

1− 53

75

)75

,

or

0.60 < p < 0.81

24. It is believed that 25% of U.S. homes have a direct satellite televisionreceiver. How large a sample is necessary to estimate the true popula-tion of homes which do with 95% confidence and within 3 percentagepoints? How large a sample is necessary if nothing is known about theproportion?

n =p̂ · q̂ · (zα/2)2

E2=

0.25 · 0.75 · (1.96)2

(0.03)2

.= 801

Assuming nothing is known about the proportion,

n =(0.5)2 · (zα/2)2

E2=

(0.5)2 · (1.96)2

(0.03)2

.= 1068

25. A recent poll showed results from 2000 professionals who interview jobapplicants. 26% of them said the biggest interview turnoff is that theapplicant did not make an effort to learn about the job or the company.A 95% confidence interval estimate was used and the margin of errorwas ±3 percentage points. Describe what is meant by the statement“the margin of error was ±3 percentage points.”

The Sampling Distribution ofSample Proportions, p̂.

When using 26% to estimate thevalue of the population percent-age, the maximum likely differencebetween 26% and the true popula-tion percentage is three percentagepoints, so the interval from 23% to29% is likely to contain the truepopulation percentage.

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KEY Statistics Test 3 Name - Tim Busken